Physics 505 Fall 2003

Transcription

1 Physics 505 Fall 2003 G. Raithel December 8, 2003 Disclaimer: The purpose of these notes is to provide you with a general list of topics that were covered in class. The notes are not a substitute for reading the textbook, nor is it guaranteed that they are complete. If you find typos, please report them to me. 9/2/2003 Units in Jackson. Gaussian and SI units were discussed. The equations for the electric field of a point charge and the magnetic field of a line current were written down for both systems. Units of charge, length, E-field, B-field, current, time, and force were discussed. Pages 78ff of the textbook were pointed out. Example. Fine-structure constant in Gauss units, α = e2 c, equals fine-structure constant in SI units, α = e2 4πɛ 0 c. The explicit calculations yielding 37 in both systems (unit-less) were sketched. The fine-structure formula for the energy levels of hydrogen-like ions was presented. The role of α 2 as a measure for the strength of EM interactions was discussed. The diagram of the lowest H-levels (n = and 2) was shown and various effects (FS, Lamb-shift) were addressed. Range of validity of Coulomb s law. Qualitative limits are given by Lamb-shift QED calculations, which indicate deviations from Coulomb s law at lengths < λ e 2π = m e c = α a 0 = 0.4 pm. A lower limit for the validity range of Coulomb s law is given by the order of the earth s radius. Math background. Some pieces of vector calculus were reviewed, including three integral theorems, parametrization of line and area integrals, coordinate transformation in volume integrals, Jacobi determinant. (see esedoglu/teaching/math-2374/example/example.html for some examples). Basic laws. Coulomb s law for force and electric field. Superposition principle. E-field of a continuous charge distribution ρ(x). Atomic electric-field unit. Reading assignment: Section about the δ-function. 2 9/4/2003 Gauss s law. Geometrical derivation using the /r 2 -dependence of the field of a point charge and the

2 superposition principle. Analytical derivation of the differential form of the law by evaluating x 4πɛ 0 ρ(x ) (x x ) x x 3 d3 x =... = ρ(x) ɛ 0. () The derivation has also provided the following insights: x x x x x x x x 2 x = (x x ) x x 3 = (x x ) x x 3 (x x ) x x 3 = 4πδ(x x ) (x x ) x x 3 = 4πδ(x x ) x x = 2 x x x = 4πδ(x x ) (2) Electrostatic potential. Brief review. Boundary conditions for field E and potential Φ on charged sheets and layers of dipoles. Poisson and Laplace equation. Derivation of Green s identities. Boundary conditions for determination of Φ(x). Dirichlet and Neumann boundary conditions. Derivation of Uniqueness theorems. 3 9/9/2003 In context with Jackson, Problem., it was shown that any curved surface can be locally parametrized using a function H(x, y ) = αx 2 + βy 2 = 2R x 2 + 2R 2 y 2. (3) Review of uniqueness theorems for Dirichlet and Neumann boundary conditions, and for a situation of n conductors with specified charges Q n on them. Integral relation from Green I: Φ(x) = 4πɛ 0 V ρ(x ) x x d3 x + 4π V The uselessness of this relation was discussed. { x x n Φ(x ) Φ(x ) n x x }da (4) 2

3 Green s function for Dirichlet Boundary Conditions (BC): G(x, x ) = 4πδ(x x ) x V G(x, x ) = 0 x V (5) is, by the uniqueness theorem, uniquely solved via F (x, x ) := G(x, x ) x x F (x, x ) = 0 x V F (x, x ) = x x x V (6) The significance of x and x (parameter / variable of PDE) was discussed. The universality of G, which only depends on the geometry, was pointed out. The generality and usefulness of the resultant solution for Φ(x) was discussed: Φ(x) = ρ(x )G(x, x )d 3 x Φ(x ) 4πɛ 0 V 4π V n G(x, x )da (7) A interpretation of G(x, x ) and of the auxiliary function F (x, x ) in terms of image charges was provided. Green s function for Neumann BC: The corresponding formalism for Neumann BC is: G(x, x ) = 4πδ(x x ) x V n G(x, x ) = 4π x V, (8) S where S is the area of V. This boundary value problem for G is, by the uniqueness theorems, solved via F (x, x ) := G(x, x ) x x F (x, x ) = 0 x V n F (x, x ) = 4π S n x x x V (9) yielding Φ(x) = ρ(x )G(x, x )d 3 x + G(x, x ) 4πɛ 0 V 4π V n Φ(x )da + Φ(x ) V (0) 3

4 with the average value of Φ on V denoted Φ(x ) V. The usefulness of this solution for Φ(x) for exterior problems with S and Φ(x ) = 0 was discussed. Electrostatic energy: Basic equations for the energy of a set of point charges, of a continuous charge distribution ρ(x) with potential Φ(x), and of a given electric field E(x) were reviewed. The self-energy contribution in the latter two was pointed out. It was noted that the energy density related to an electric field leads to an electrostatic pressure, p = σ2 2ɛ 0 ˆn, on charged conductor surfaces. Using the uniqueness theorem for charged conductors, it was shown that a set of n conductors with potentials V i and charges Q i satisfies the capacitor equation V j = n i= p j,iq i and its inverse, Q i = n j= C i,jv j. The reasoning for the nomenclature for the capacitances C i,i and the induction coefficients C i,j with i j was explained. Expressions for the energy were given: W = 2 n p i,j Q i Q j = 2 i,j= n C i,j V i V j () A brief review of how to calculate the capacitance of a conductor pair was provided (assume charges ±Q, find E(x), calculate V = 2 E(x) dl, then C = Q/V ). i,j= 4 9//2003 Some questions concerning the proof of the capacitor equation, V i = n j= p i,jq j, which is the only non-trivial part of the discussion on page 43 of the textbook, have been clarified. Variational principles. It has been shown that the functional I[ψ] = ψ 2 d 3 ρ(x) x ψd 3 x (2) 2 ɛ 0 V becomes minimal if the test function ψ(x) satisfies the Poission Equation and Dirichlet BC. The benefit is two-fold: firstly, it is seen that a variational principle exists that leads to the Poission Equation. Secondly, it follows that variational methods can be used to find approximate solutions to the electrostatic boundary-value problem. To see this, consider a given set of test functions {ψ α,β,... (x)} satisfying the Dirichlet BC of the problem, with parameters α, β,... Then, within the chosen set of test functions the function ψ α0,β 0,...(x) for which I[ψ] becomes minimal represents the best approximation to the actual solution Ψ(x). To identify ψ α0,β 0,...(x), calculate I(α, β,...) := I[ψ α,β,... ] and find the parameters (α 0, β 0,...) for which α I(α, β,...) = 0, β I(α, β,...) = 0 etc., and for which the value of I is lowest. A functional suitable for Neumann BC has been given. The class has been advised to read the examples provided in the textbook. Relaxation methods. For charge-free volumes with Dirichlet BC, the cross-averaging method and its foundation in the previously discussed variational principle has been explained. The Jacobi and the Gauss-Seidel iteration methods have been shown. A hyper-relaxation method yielding significantly improved convergence has been presented. The accuracy of cross-average relaxation method in terms of the grid size has been derived. Improved methods that use weighted cross- and square-averages and a method that allows for the incorporation of non-zero charge densities have been pointed out. V 4

5 5 9/6/2003 The connection between the image charge method and the Dirichlet Green s function G D has been explained. Assume that a charge q at location x in the volume of interest produces image charges q i (x ) at locations x i outside the volume of interest. Since the image charges are proportional to q, we may write q i (x ) = q q i (x ), with the normalized image charges q i (x ) measuring the values of the image charges relative to q. Then, G D (x, x ) = x x + F D(x, x ) = x x + i q i (x ) x x i (x ) (3) The equations and boundary conditions for G D and F D are: G(x, x ) = 4πδ(x x ) x V G(x, x ) = 0 x V (4) and F (x, x ) = 0 x V F (x, x ) = x x x V. (5) These equations make it evident that F D plays the role of the potential of image charges generated by a charge 4πɛ 0. An important consequence of the connection between Green s function and image charges is that once a Dirichlet problem has been solved with the image charge method, its Green s function is also known, and a larger class of problems can be solved. Example. The case of a conducting plane at x = 0 and the volume of interest being the half-space x > 0 has been discussed. A charge q at location x produces one image charge q = q at location x (x ) = x (2x ˆx)ˆx. The normalized value of the image charge is, and consequently G D (x, x ) = x x x x + (2x ˆx)ˆx. (6) According to the Green s function formalism for Dirichlet BC, it is Φ(x) = 4πɛ 0 + 4π volume x >0 plane x =0 { ρ(x ) x x Φ(0, y, z ) x { x x + (2x ˆx)ˆx } dx dy dz x x x x + (2x ˆx)ˆx } dy dz (7) Note that in this problem n = x, and that x is the cartesian x-coordinate of x (not the radial coordinate of x, as in numerous other equations in the textbook). The first line in Eq. 7 represents the term that immediately follows from the image charge method and the superposition principle. The second line requires knowledge of the Green s function formalism, and 5

6 allows one to treat the extended class of problems in which an arbitrary potential on the surface V is specified. The image-charge problem of a charge q outside a grounded, conducting sphere with radius a has been discussed. We have obtained the potential Φ(x), the induced surface charge density σ on the sphere, the total induced charge, and the forces on the charge and the sphere. The superposition principle allows for some straightforward extensions. These include the cases of a charge q outside a conducting sphere with a given potential V or with a given total charge Q. The basic method is explained in the following for fixed V. We identify quantities obtained for the case of a charge q outside the grounded sphere with a subscript I. Consider the solutions of the following problems: I=grounded sphere with charge q outside. II=sphere on potential V and no charge outside the sphere. The solution of case II is a constant potential V on and inside the sphere, and Φ II (x) = V a x outside the sphere, with a total charge of 4πɛ 0 V a evenly distributed over the surface of the sphere. The sum of the charge densities of cases I and II, and the sum of the potentials, Φ = Φ I + Φ II, satisfy the boundary conditions. That is: the sums produce the correct external charge distribution and the correct potential on the boundary, respectively. Also, due to the superposition principle, the sum potential and the sum charge distribution are a solution of the Poisson equation. Due to the uniqueness theorem, this must be the only solution for the given surface potential V and the given charge distribution in the volume of interest. Thus, outside the sphere it is Φ = Φ I + Φ II = Φ I + V a x. The surface charge density on the surface is σ = σ I + σ II = σ I + ɛ 0V a, and the total charge is q + 4πɛ 0V a. Since the electric fields also follow the superposition principle, the force is F = F I + qv a y ŷ. 2 Sometimes, simple tricks allow for the treatment of seemingly unrelated problems. In the present instance, the problem of a charge q outside a grounded, conducting sphere can be easily twisted in a way that allows for a simple calculation of the potential and the charge densities of a conducting sphere in a homogeneous electric field. Read textbook. The Dirichlet Green s function of a spherical surface has been deduced form the corresponding imagecharge problem. In a system of spherical coordinates with z-axis pointing to the location of interest x, and denoting the angle between the vectors x and x by γ, the Green s function is seen to be G D (x, x ) = G D (x, x, cos γ) = x2 + x 2 2xx cos γ x 2 x 2 a 2 + a 2 2xx cos γ, (8) which is symmetric (as must be), and on the surface it is n G D(x, x ) x =a = x G D(x, x x 2 a 2, cos γ) x =a = a x 2 + a 2 2ax cos γ 3. (9) Noting that the spherical coordinates of the source location in the chosen frame are (x, γ, φ ), for the calculation of the potential one writes 6

7 Φ(x) = 4πɛ 0 + 4π volume x >a surface x =a ρ(x, γ, φ )G D (x, x, cos γ)x 2 dx dφ d cos γ Φ(a, γ, φ a(x 2 a 2 ) ) x2 + a 2 2ax cos γ 3 dφ d cos γ. (20) It was shown that cos γ = cos θ cos θ + sin θ sin θ cos(φ φ ). To disentangle the reference frame from the point of interest x, one can then write Φ(x) = 4πɛ 0 + 4π volume x >a surface x =a ρ(x, θ, φ )G D (x, x, cos γ(θ, θ, φ, φ ))x 2 dx d cos θ dφ Φ(a, θ, φ a(x 2 a 2 ) ) x2 + a 2 2ax [cos θ cos θ + sin θ sin θ cos(φ φ )] 3 dφ d cos θ (2) The class has been advised to read the example of two hemispheres on opposite potentials discussed in the textbook. 6 9/8/2003 Basic concepts of electrostatics have been reviewed: Uniqueness theorems, Green s function, relations between Green s function and image charges. The problem of two hemispheres with radius a on potentials V and V in a charge-free space has been briefly discussed. The resultant potential follows from the surface part of Eq. 2, and is Φ(x) = V a(x2 a 2 ) 4π 2π φ =0 dφ cos θ = { } d cos θ x2 + a 2 2ax cos γ 3 x2 + a 2 + 2ax cos γ 3, where cos γ = cos θ cos θ +sin θ sin θ cos(φ φ ). The integral can be calculated along the +z-axis, where γ = θ. Two methods of how to obtain an approximate solution for x a, valid for all θ, have been pointed out: ) The exact potential can be expanded along the +z-axis, yielding an expression of the form Φ(z) = n=,3,... a nz n with expansion coefficients a n. In the given problem, the coefficients a n for even n vanish because of symmetry. The general potential is then given by Φ(x, θ) = n=,3,... a nx n P n (cos θ). This method will be further discussed later. 2) The square root under the integral may be expanded in terms of a small parameter ɛ that involves cos θ. One may write, for instance, (x 2 + a 2 2ax cos γ) 3/2 = (x 2 + a 2 ) 3/2 ( ɛ) 3/2 with ɛ = 7 (22) 2ax cos γ x 2 + a 2 (23)

8 The expansion in ɛ leads to solvable integrals, and, eventually, to a power series involving odd powers of cos θ. The general strategy of solving the Laplace equation using the variable separation method has been outlined. Usually, there are three main steps involved: ) Find a complete set of orthogonal functions that solve the Laplace equation and fit the given geometry and the BC. This can often be accomplished using the method of variable separation, which usually involves the following steps: a) Consider the symmetry of problem to make the best choice of a coordinate system. Box problems and other problems involving (mostly) right angles and straight surfaces are treated with cartesian coordinates. Problems with circles, circle segments, cylinders etc. are usually treated in cylindrical coordinates, and problems involving spheres or sections of spheres with spherical coordinates. b) Write Φ = 0 in the coordinates identified in step a). c) Write down a set of solutions obtained from the variable separation method. Completeness may matter. d) Elimination process. To simplify the further steps, use simple boundary conditions to reduce the number of basis functions and expansion coefficients from the result of c). Simple BC are, for instance, surface portions that are entirely on zero potential. Often, the formation of certain linear combinations helps (see, for instance, the sinh(γz) term in Eq. 2.53). Eliminate diverging terms as appropriate. 2) Write the potential as a linear superposition of the functions left over after step d). 3) Find the expansion coefficients by surface integrals over surfaces with known boundary values. Use the orthogonality of the complete set of functions. As a specific example, a rectangular three-dimensional box with five faces on zero potential and the upper xy-face on a potential V (x, y) was discussed. This problem is treated using cartesian coordinates. The proper use of sin, cos, exp, sinh and cosh for the basis functions was explained. It was explained why the superposition principle allows one to generalize the result to arbitrary Dirichlet BC on the box (all sides on arbitrary potential, not just one). As another example, the two-dimensional Laplace equation has been separated in cylindrical coordinates, and proper basis solutions for the case of a two-dimensional grounded corner or edge have been obtained. In addition to your regular reading of the textbook, make sure you take special care of the following Reading assignments: Refresh undergraduate knowledge of the separation method. If necessary, consult an undergraduate textbook such as Griffiths, Introduction to Electrodynamics. Read section 2.8 in the textbook (about orthogonal function sets). Review the orthonormality and the completeness properties, and the given examples (discrete Fourier series, Fourier transform...). Read the portions of sections that were skipped in class. 8

9 7 9/25/2003 The separation method has been reviewed. In connection with Chapter 2.0 of the textbook, the relation between the Laplace Equation and analytic functions has been discussed. Proofs for the following facts have been outlined: For any analytic function f(z) = u(x, y) + iv(x, y), with z = x + iy, the Cauchy-Rieman equations u(x,y) apply: x = v(x,y) y and v(x,y) x = u(x,y) y. u = v = 0. Therefore, the imaginary and reals parts of analytic functions often coincide with the solutions of 2D potential problems. More examples were discussed: The separation method has been used to treat the 2D potential problem of corners and edges with straight surfaces on a constant potential V 0 and a potential V (φ) on the cylindrical section of the surface. The solution is Φ(ρ, φ) = n= a n ρ nπ β ( ) nπ sin β φ with a n = 2 βr nπ β β φ=0 ( ) nπ Ṽ (φ) sin β φ dφ (24) V 0 ( ) V 0 R V( ) V 0 = + V 0 V 0 V 0 ~ V( )=V( )-V0 0 ~ ( )= ( )-V 0 0 Figure : 2D problem of a corner on potential V 0 on the straight sections of the surface and potential V (φ) on the cylindrical section. The figure shows how to use the superposition principle to reduce the problem to the case of zero potential on the straight surfaces. As example illustrating the use of the results, we have estimated the outcome of a molecular-beam deflection experiment in which molecules with a linear polarizability are deflected when traveling parallel to the edge of a thin, grounded conducting plate surrounded by a cylinder on a potential V. Two concentric cylinders with radii a < b and surface potentials V a (φ) and V b (φ) on the surfaces. Discussion of the general solution Φ(ρ, φ) = a 0 + b 0 ln ρ + ± n=±,±2,... (a n ρ n + b n ρ n ) (A n cos(nφ) + B n sin(nφ)), (25) which is analogous to Eq. 2.7 in the textbook. (issues: which terms to drop in what case, how to find the remaining coefficients). Finite element method. The 2D case of finite-element functions of pyramidal shape that are arranged on a square grid has been discussed. Reading: improved method using triangular shape functions. 9

10 8 9/30/2003 Solution of Φ = 0 by separation of variables in spherical coordinates: It has been outlined how the φ-, the r- and the θ-equations are obtained and solved. The solution of the θ-equation has been limited, for now, to the case of azimuthal symmetry (m = 0 all over). One may add that for the φ- and r-equations and for given values of the separation variables m 2 and l(l + ) two linearly independent solutions contribute, which is the maximum number possible because the underlying linear homogeneous differential equations are of 2nd order. For the θ-equation only one solution P l (cos θ) is considered as physically relevant solution. The other solution, which formally exists, is obtained by making the other possible choice of α (textbook, Eqn. 3.3, st line). That other solution diverges at x = cos θ = ± and is therefore discarded. Reading: Properties of Legendre polynomials. Orthogonality: P l(x)p l (x) dx = 2 2l+ δ l l Function Expansions: From orthogonality and completeness it follows that for well-behaved f(x) on [, ] it is f(x) = l=0,,.. A lp l (x) with A l = 2l+ 2 P l(x)f(x) dx. Closure: It has been explained how to obtain the closure relation δ(x x ) = l=0,,.. 2l+ 2 P l(x)p l (x ). The use of recursion relations between P l s has been briefly discussed. Potential boundary value problems with azimuthal symmetry: The potential is of the general form Φ(r, θ) = l=0,,.. (A lr l + B l r l )P l (cos θ), where the A l and B l are determined from BC. Example: For given potential V (θ) on a sphere with radius a, the interior solution has B l = 0 and A l = 2l+ 2a l V (θ)p l(cos θ) d cos θ. (I believe I had the factor 2l+ 2 up-side-down when I wrote this on the board - if so please correct in your notes). For the exterior solution, A l = 0 and B l = (2l+)al+ 2 V (θ)p l(cos θ) d cos θ. Another Method of obtaining the expansion coefficients A l and B l is to perform an exact calculation of Φ(z) along the z-axis, where P l. Then, expand Φ(z) in either negative or positive powers of z, dependent on whether you seek an exterior or interior solution, respectively. For an interior solution, you may find, for instance, Φ(z) = l=0,,.. A lz l. If the problem has azimuthal symmetry, then the potential at a general location is Φ(r, θ) = l=0,,.. A lr l P l (cos θ), with the same coefficients A l as in the expansion of Φ along the z-axis. The reason why the method works is the uniqueness of the expansion of both Φ(z) along the z-axis and Φ(r, θ). As an application, we have derived the following important expansion of x x : x x = r< l r l+ l=0,,... > P l (cos θ), where r < = min(r, r ) and r > = max(r, r ) (26) As an application of the expansion Eq. 26, we have expanded the potential of a ring charge Q that is concentric with the z-axis, has a radius a, and has a height b above the xy-plane. This example is covered in the textbook on page 03. What s most important to note about this example is that the use of the expansion Eq. 26 saves us from having to expand z as an infinite power series of z: 2 +c 2 2cz cos α The hard way to find Φ(r, θ) would be to first write Φ(z, θ = 0) = q 4πɛ 0 z, to then 2 +c 2 2cz cos α 0

11 expand this either in powers of z for small z or in powers of z for large z, to write the result in the form Φ(z) = l=0,,.. A lz l or Φ(z) = l=0,,.. B lz l, and to write the general potential as Φ(r, θ) = l=0,,.. A lr l P l (cos θ) or Φ(r, θ) = l=0,,.. B lr l P l (cos θ). It will finally be left to show that the coefficients can be written in the concise form A l P c l+ l (cos α) and B l = c l P l (cos α), respectively. Easy way: The use of Eq. 26 yields, after a few lines of just writing down facts rather than doing calculations, Φ(r, θ) = Q 4πɛ 0 l=0,,... r l < r l+ > P l (cos α)p l (cos θ), where r < = min(r, c) and r > = max(r, c). 9 0/02/2003 To illustrate the use of the expansion x x = l=0 r l < r l+ > P l (cos θ), the problem of a ring with radius a, charge Q and center at ẑb has been discussed (see page 03 of textbook). By direct expansion of the exact potential Φ(z) along the z-axis written as a closed square-root expression, we have obtained the terms l = 0,, 2 in the exterior case z > c and seen that the resultant power series has coefficients involving P l (cos α). It was then shown that an application of the expansion x x = l=0 r l < r l+ > P l (cos γ) for the case x = ẑz and x = ˆρa + ẑb immediately leads to the expansion of the potential up to arbitrary l, valid for arbitrary x. Cones and tips with azimuthal symmetry. It was outlined how to solve the Legendre differential equation for Dirichlet BC P (cos β) = 0, with 0 < β < π. The solutions P ν (cos θ) are called the Legendre functions of the st kind of (generally non-integer) order ν, and they are equivalent to hypergeometric functions 2 F ( ν; ν + ; ; cos θ). Some properties of the P ν (cos θ) were pointed out. For given β, a countable set of such functions {P νk (cos θ) k =, 2,...} satisfies the BC P ν (cos β) = 0. The significance of the counting index k is that P νk (cos θ) has its k-th zero at cos β. The set {P νk (cos θ) k =, 2,...} is a complete and orthogonal set of functions on the interval [cos β, ] (with f(cos β) = 0). Thus, the general potential near tips and cones is of the form Φ(x) = A k r ν k P νk (cos θ) (27) k= Reading assignment: Properties of the leading term k = of that expansion (textbook p. Criticality of the lowest order ν (β). Qualitative change of field behavior at β = π/2. 06f). Potential expansion in spherical coordinates - general case without azimuthal symmetry. Φ(x) = l l=0 m= ( Alm r l + B lm r l ) Y lm (θ, φ) (28) Some properties of the spherical harmonics Y lm (θ, φ) were reviewed, including orthogonality and closure relations. The method of how to determine the expansion coefficients in the case of Dirichlet BC on a sphere with radius r = a was briefly reviewed. Reading assignment: Chapter 3.5 of the textbook (properties of spherical harmonics). The following important expansion of the free-space Green s function has been pointed out:

12 x x = r< l r> l+ P l (cos γ) = l=0 l l=0 m= 4π 2l + r< l r> l+ Y lm (θ, φ)ylm(θ, φ ) (29) The separation of the Laplace equation in cylindrical coordinates was explained. Two cases of physical significance exist: Case : We consider the case that the potential is zero on a cylinder mantle of radius a. Then, the solutions are of the form Φ(ρ, z, φ) = exp(±iνφ) exp(±kz)ω ν (kρ), and linear combinations thereof. The index ν usually is 0,, 2..., and the variable k is real and 0. The radial function Ω ν (kρ) is a Bessel function (a solution of the Bessel differential equation). The properties - asymptotic behavior, roots, etc. - of the Bessel functions J ν (x), J ν (x), the Neumann functions N ν (x), and the Hankel functions H ν () (x) = J ν (x) + in ν (x) and H { ν (2) (x) = J ν (x) } in ν (x) were discussed. The pair {J ν (x), N ν (x)} is always independent, and so is H ν () (x), H ν (2) (x). Most importantly, the function set { ρ Jν (ρ x νn a ) ν and fixed, and n =, 2, 3... } (30) is a complete orthogonal set on the interval [0, a] with BC f(ρ) = 0 at ρ = a. Thereby, x νn stands for the n-th root of J ν (x) = 0. Note that each value of ν generates a complete orthogonal set. Using the orthogonality relation a 0 ρj ν (ρ x νn a )J ν(ρ xνn a2 )dρ = a 2 J ν+(x 2 νn )δ n,n (3) any function f(ρ) that vanishes at ρ = a can be expanded as f(ρ) = n= A νnj ν (ρ x νn ) with coefficients A νn = 2 a 2 J 2 ν+ (x νn) a 0 ρf(ρ)j ν (ρ x νn )dρ (32) a These findings can be used to expand the potential in cylindrical volumes of radius a with V = 0 on the cylinder mantle. If the cylinder is closed on both top and bottom, we can also - by the superposition principle - require that only the top or the bottom lid is on a non-zero potential. To be specific, assume a bottom lid with V = 0 at z = 0 and a top lid at z = L with potential V (ρ, φ). Convince yourself by a calculation that the potential in the cylinder can be expanded as a Φ(x) = J m (k mn ρ) sinh(k mn z) (A mn sin(mφ) + B mn cos(mφ)) (33) m=0 n= with k mn = x mn a and 2

13 A mn = B mn = 2 πa 2 J m+ (k mn a) sinh(k mn L) 2 πa 2 J m+ (k mn a) sinh(k mn L) 2π a 0 0 2π a 0 0 ρv (ρ, φ)j m (k mn ρ) sin(mφ)dρ, n =, 2,... ρv (ρ, φ)j m (k mn ρ) cos(mφ)dρ {, n =, 2,... /2, n = 0 (34) Case 2: The other case of significance is that the potential is zero on the top and the bottom of the cylinder volume, while on the mantle of radius a the potential is non-zero and equal to V (z, φ). Assuming that the bottom of the cylinder is located in the plane z = 0 and the top in the plane z = L, the interior solutions are then of the form exp(±iνφ) sin(kz)ω ν (kρ). There, Ω ν (kρ) is a modified Bessel function, and the value of k satisfies a quantization condition ka = nπ with integer n. The commonly used modified Bessel functions are I ν (x) = i ν J ν (ix) and K ν (x) = π 2 iν+ H ν () (ix). They are both real and linearly independent. I ν (x) is regular at x = 0 and diverging for x, while K ν (x) is regular for x, and diverging for x 0. If there is no radial restriction to the volume of interest, expansions of the potential in non-countable function sets characterized by one or more continuous indices are useful. The potential can, for instance, be expanded into a Fourier integral. In cases of cylindrically symmetric BC without radial restriction we can use the complete orthogonal set of functions { ρjm (kρ) k real and k 0 }, which has the orthogonality relation 0 ρj m (kρ)j m (k ρ)dρ = k δ(k k ) m. (35) Thus, in the (charge-free) volume z > 0 and boundary conditions Φ(ρ, z = 0, φ) = V (ρ, φ) the potential is given by a Fourier-Bessel integral Φ(ρ, z, φ) = m=0 0 where the coefficient functions are dk exp( kz) J m (kρ) (A m (k) sin(mφ) + B m cos(mφ)), (36) A m (k) = k π B m (k) = k π 2π ρ=0 φ=0 2π ρ=0 φ=0 ρv (ρ, φ)j m (kρ) sin(mφ)dρdφ, m =, 2,... ρv (ρ, φ)j m (kρ) cos(mφ)dρdφ {, m =, 2,... /2, m = 0 (37) Reading: Similar methods involving spherical Bessel functions (p9 of textbook). Good exercise: Derive completeness relations for complete orthogonal sets (CONS) of Bessel functions on a finite interval [0, a], on [0, [, and for CONS of spherical Bessel functions on [0, [. 3

14 Expansion of Green s functions using complete sets of orthogonal functions. The general usefulness of the Green s function has already been discussed earlier. The purpose of the following is to extend the application of expansion methods from mere solutions of the Laplace equation to the Green s function. Previous example: The free-space Green s function can be expanded in spherical harmonics as G(x, x ) = x x = l l=0 m= 4π 2l + r< l r> l+ Y lm (θ, φ)y lm(θ, φ ). (38) This result is important in the multipole expansion (Chapter 4 of the textbook). It can also be used to expand the Green s functions of problems that can be treated with the image charge method. Example: Dirichlet Green s function for the exterior volume of a sphere with radius a. For the term in the Green s function that corresponds to the image charge we use the following: Relative image charge size = a r with r being the radial coordinate of the source location x. r > = r, where r is the radial coordinate of the observation coordinate x. r < = a2 r Inserting these figures it is found: G(x, x ) = l l=0 m= ( 4π r 2l + Y lm(θ, φ)ylm(θ, φ l ) < r> l+ ) a ( a2 rr )l+. (39) 0 0/07/2003 Systematic methods of how to expand Green s functions. As an example, we consider the Green s function of the volume between two concentric shells with radii a < b. This example is lengthier than most homework or exam problems of this kind. Step : Write down the differential equation for the Green s function with δ-function in spherical coordinates, G(x, x ) = 4πδ(x x ) = 4π δ(r r ) r 2 δ(φ φ )δ(cos θ cos θ ) Step 2: On right side, use completeness relations for two out of the three involved δ-functions. In the present case of spherical coordinates, we only have completeness relations for the angular coordinates. Therefore, we write G(x, x ) = 4π δ(r r ) r 2 l l=0 m= l 4 Y lm(θ, φ )Y lm (θ, φ)

15 and re-sort: G(x, x ) = l l=0 m= l { 4π δ(r } r ) r 2 Ylm(θ, φ ) Y lm (θ, φ) (40) Note: In cylindrical or cartesian coordinates, completeness relations are often known for all three of the involved δ-functions, and a choice must be made. Depending on that choice, different but equally valid expansions of the Green s function are obtained (see homework problems). Step 3: On left side, expand the Green s function using the orthogonal function sets that have also been used in Step 2. Note that x only enters as a parameter of the calculation; acts on x. G(x, x ) = l l=0 m= l A lm (r r, θ, φ ) Y lm (θ, φ) Step 4: Write Laplacian in the proper coordinates and take derivatives of all orthogonal functions that have been introduced in Steps 2 and 3. It many cases you will have to use a known differential equation to take these derivatives. Typically, you will have to use the (plain or generalized) Legendre differential equation to eliminate angular derivatives of spherical harmonics or Legendre functions. To eliminate radial derivatives of Bessel functions, employ the Bessel differential equation. In the present example, we use the generalized Legendre differential equation. G(x, x ) = Using x = cos θ, it is G(x, x ) = l l=0 m= l [ r l l=0 m= l 2 r 2 r + r 2 sin θ [ r 2 r 2 r + r 2 θ sin θ θ + r 2 sin 2 θ x ( x2 ) x 2 ] φ 2 A lm (r r, θ, φ ) Y lm (θ, φ) ] m2 r 2 sin 2 A lm (r r, θ, φ ) Y lm (θ, φ) θ d The [ generalized ] Legendre differential equation allows [ us to write ] dx ( x2 ) d dx P l m (x) = l(l + ) + m2 x P m 2 l (x), and also x ( x2 ) x Y lm = l(l + ) + m2 x Y 2 lm. Thus, with x 2 = sin 2 θ G(x, x ) = l l=0 m= l G(x, x ) = [ r 2 l(l + ) r r2 r 2 + m2 r 2 sin 2 θ l l=0 m= l {[ r d 2 l(l + ) r dr2 r 2 ] ] m2 r 2 sin 2 A lm (r r, θ, φ ) Y lm (θ, φ) θ } A lm (r r, θ, φ ) Y lm (θ, φ) (4) Step 5: Expansions in orthogonal sets of functions are unique. Thus, we can separately equate the coefficients of the Y lm in Eq. 40 and Eq. 4. Dividing the resultant equation by Ylm (θ, φ ), we find [ d 2 l(l + ) r r dr2 r 2 ] { Alm (r r, θ, φ } ) Y lm (θ, φ ) 5 = 4π δ(r r ) r 2

16 Step 6: Noticing that the expression in the curly brackets of the last equation can possibly only depend l, r and r, we define the reduced (radial) Green s function g l (r, r ) = A lm(r r, θ, φ ) Y lm (θ, φ ). Note that the radial Green s function only depends on one of the two indices of the orthogonal function set employed in Steps 2 and 3. This is an exception. In most other cases, reduced Green s functions will depend on all indices of the utilized orthogonal function sets. We proceed to solve [ d 2 ] l(l + ) r r dr2 r 2 g l (r, r ) = 4π δ(r r ) r 2 (42) Outside the location of the δ-function inhomogeneity, the solutions are of the form g l (r, r ) = Ar l Br l. More specifically, to match the boundary condition g l (r = a, r ) = 0, for r < r it must be g l (r, r ) r l a2l+. To match the boundary condition g r l+ l (r = b, r ) = 0, for r > r it must be g l (r, r ) r l rl. b 2l+ Further, g l (r, r ) must be symmetric in r and r, and it must be continuous at r = r. To satisfy all these conditions, g l (r, r ) must be of the form g l (r, r ) = C (r l< a2l+ r l+ < ) ( ) r> l rl > b 2l+, with a constant C to be determined to match the behavior near the δ-function inhomogeneity. Also, r < = min(r, r ) and r > = max(r, r ). Step 7: Find C. Integrating Eq. 42 from r ɛ to r + ɛ with ɛ 0 and dropping vanishing terms we find A lengthy but simple calculation yields Step 8: d dr (rg l(r, r )) r=r +ɛ d dr (rg l(r, r )) r=r ɛ = 4π r 4π C = ( (2l + ) ( ) ) a 2l+ b Going backward through all steps, the Green s function expansion is assembled into the final result: g l (r, r 4π ) = ( (2l + ) ( ) ) a 2l+ b (r l< a2l+ r l+ < ) ( ) r> l rl > b 2l+, A lm (r r, θ, φ 4π ) = ( (2l + ) ( ) ) a 2l+ b (r l< a2l+ r l+ < ) ( ) r> l rl > b 2l+ Ylm(θ, φ ) 6

17 and finally G(x, x ) = 4π l l=0 m= l Y lm (θ, φ ) Y lm (θ, φ) ( (2l + ) ( ) ) a 2l+ b (r l< a2l+ r l+ < ) ( ) r> l+ rl > b 2l+ Reading: Use of the above Green s function to calculate the potential produced by charge distributions (Chapter 3.0 of the textbook). Reading: Application of the expansion method to expand the free-space Green s function in cylindrical coordinates (Chapter 3. of the textbook). Eigenfunction expansion of the Green s function. General method. The solutions ψ(x) of the eigenvalue problem ( + f(x) + λ)ψ(x) = 0 with eigenvalues λ can be arranged to form an orthonormal, complete set of functions on the definition rage of the equation. Boundary conditions may apply. Closed volumes lead to countable sets {ψ n (x), n =, 2,..} with corresponding eigenvalues λ n, while open volumes have sets with at { least one continuous parameter. } In free space and f(x) = 0, for instance, one may choose ψ k (x) = 3 exp(ik x), k R 3 with eigenvalues λ k = k 2. 2π The analogy of the situation to quantum mechanics has been discussed. In that language, + f(x) is a Hermitian linear differential operator. As is well known, such operators plus their boundary conditions generate orthonormal, complete sets of eigenfunctions. It has been shown that orthogonality and completeness of the set {ψ n (x)} with eigenvalues {λ n } lead to an expansion of the Green s function. Assume ( + f(x) + λ n )ψ n (x) = 0, with ψ n (x) satisfying any boundary conditions that may apply. The Green s function is defined as the solution of ( + f(x) + λ)g(x, x ) = 4πδ(x x ), with G(x, x ) satisfying the same boundary conditions as the ψ n (x). It follows G λ (x, x ) = 4π n ψ n(x ) ψ n (x) λ n λ. In the special context of the Laplace equation, set f(x) = 0 and λ = 0 in the above equations. Then, for discrete sets of eigenfunctions satisfying ( + λ n )ψ n (x) = 0, n =, 2,.., boundary conditions, and orthonormality ψn(x) ψ n (x)d 3 x = δ n,n we have 7

18 G(x, x ) = 4π n ψ n(x ) ψ n (x) λ n. For sets of eigenfunctions with continuous eigenvalues satisfying ( + λ k )ψ k (x) = 0, continuous k, boundary conditions (if any), and orthonormality ψ k (x) ψ k (x)d3 x = δ(k k ) we have G(x, x ψk ) = 4π (x ) ψ k (x) dk. k λ k Sometimes there is more than one continuously variable index for the eigenfunctions. Also, combinations of continuous and discrete indices exist. In all these cases, add integrals or sums in the above equations as appropriate. Two examples have been briefly discussed (rectangular box and free space). 0/6/2003 Multipole expansion. The first case in which such expansions are useful is if we are concerned with the potential generated by a small charge distribution in an otherwise charge- and field-free space. We consider the potential of a charge distribution ρ(x ) of typical extension R at locations x with x = r > R. Then, application of the free-space Green s function expansion Eq. 38 yields Φ(x) = 4πɛ 0 with spherical multipole moments l l=0 m= l 4π 2l + q lm r l+ Y lm(θ, φ) (43) q lm = ρ(x ) r l Y lm(θ, φ ) d 3 x (44) Alternately, an expansion of x x in cartesian coordinates for fixed x around x = 0 yields Φ(x) = 4πɛ 0 l=0 l k, m, n = 0 k + m + n = l [ k!m!n! k x k m y m n ( z n x x )] x =0 ρ(x )x k y m z n dx dy dz = 4πɛ 0 l=0 l k, m, n = 0 k + m + n = l P (l) kmn r 2l+ (x, y, z, r) Q (l) kmn (45) 8

19 Thereby, l identifies the order of the expansion terms (as in the expansion in spherical coordinates). P (l) kmn is a l-th order polynomial in the observation point coordinates (x, y, z, r = x 2 + y 2 + z 2 ), leading to an overall radial dependence of the l-th order terms (as in the expansion in spherical coordinates). r l+ Further, Q (l) kmn is the l-th cartesian multipole moment Q (l) kmn = ρ(x )x k y m z n dx dy dz (46) The monopole (l = 0) and dipole (l = ) terms have been stated. It has been shown in class that after resorting of terms the quadrupole (l = 2) contribution takes the form given in Eq. 4.0 of the textbook, with the quadrupole tensor given in Eq Unless noted otherwise, Eq. 4.9 and 4.0 are used to calculate cartesian quadrupole moments and their potentials and fields. To further demonstrate the systematics in the cartesian multipole expansion, in the following the next terms are given. You may check that the term l = 3 is Φ 3 (x) = 4πɛ 0 3 i, j, k = i j k [ A ijk x i x j x k ( ) ] x x x =0 ρ(x )x ix jx k d 3 x = 4πɛ 0 3 i, j, k = i j k 5x i x j x k 3r 2 (δ ij x k + δ jk x i + δ ki x j ) A ijk r 7 Qijk where the indices i, j, k refer to the three components of cartesian coordinates, Qijk is the octupole moment Q ijk = ρ(x )x i x j x k d3 x, and A ijk = if all indices are different, A ijk = 2 if exactly two indices are equal and A ijk = 6 if all indices are equal. The general form of this result conforms with the l = 3-term of Eq. 45; note, however, the different meanings of the indices. It is, for instance, Q 22 = Q (3) 20. Similarly, you should find for l = 4 Φ 4 (x) = = 4πɛ 0 4πɛ 0 3 i, j, k, m = i j k m 3 i, j, k, m = i j k m [ A ijkm x i x j x k x m H ijkm (x) A hijk r 9 Qijkm with ( ) ] x x x =0 ρ(x )x ix jx kx m d 3 x H ijkm (x) = 05x i x j x k x m 5r 2 (δ ij x k x m + δ jk x i x m + δ ik x j x m + δ im x j x k + δ jm x i x k + δ km x i x j ) + 3r 4 (δ ij δ km + δ jk δ im + δ ik δ jm ) (47) There, Qijkm = ρ(x )x i x j x k x m d 3 x, and A ijkm = 2 if exactly two indices are equal, A ijkm = 6 if exactly three indices are equal, and A ijkm = 24 if all indices are equal. This result is of the same form 9

20 as the l = 4-term in Eq. 45. Again, note the different meanings of the indices. It is, for instance, Q 23 = Q (4) 2. Important fact a: Out of the (l + 2)(l + )/2 cartesian multipole moments Q (l) kmn of order l, defined in Eq. 46, there are only 2l + independent ones. This number equals the number of spherical moments of order l; the latter are all independent from each other. Important fact b: The lowest-order non-vanishing multipole moments are invariant under translations of the origin (not under rotations). For the cartesian moments, this fact follows from a transformation rule obtained in the following. Assume that the multipole moments in a certain frame are labeled Q (l) kmn. The origin is then translated by (l) x 0 = (x 0, y 0, z 0 ). Then, the transformed moments ˆQ kmn are ˆQ (l) kmn = = = = k ρ(x)(x x 0 ) k (y y 0 ) m (z z 0 ) n dxdydz m n ( ) ( k m a b a=0 b=0 c=0 k m n ( ) ( k m a b a=0 b=0 c=0 k m n a=0 b=0 c=0 a+b+c<l ) ( ) n ( x c 0 ) k a ( y 0 ) m b ( z 0 ) n c ρ(x)x a y b z c dxdydz ( k a ) ( n c ) ) ( m b ( x 0 ) k a ( y 0 ) m b ( z 0 ) n c Q (a+b+c) abc ) ( n c ) ( x 0 ) k a ( y 0 ) m b ( z 0 ) n c Q (a+b+c) abc + Q(l) kmn (48) There, the ( ) are binomial coefficients. Since the expression in the curly brackets only contains multipole moments of orders lower than l, it is shown that the lowest-order non-vanishing multipole moments are independent of translations of the origin. ( Important fact c: The contributions of the multipoles to the potential scale as R ) l, r r where R is the size of the source and the origin is assumed to be chosen inside the source (which is a prerequisite for the expansion to make sense). This scaling behavior explains why the multipole expansion is a powerful tool. Review the following: Electric fields associated with multipole moments. Potentials and fields of an electric dipole. Distinction and relation between idealized multipoles and their approximate realizations. matter has been discussed in class for an electric dipole. The 20

21 In preparation of future subjects, the following results have been obtained. Consider the integral over the electric field inside a sphere with radius R. If the sphere includes all charges ρ(x), then r<r E(x)d 3 x = p with dipole moment p = xρ(x)d 3 x. (49) 3ɛ 0 r<r If all charges are outside the sphere, then the average electric field in the sphere equals the field at the center, i.e. r<r Reading: detailed derivation of these equations in Chapter 4.. E(x)d 3 x = 4π 3 R3 E(0). (50) The second case in which multipole expansions are useful is if one is concerned with the energy of a charge distribution ρ(x) of typical size R in an external electric field that varies over length scales much larger than R. We assume that the shape of the distribution ρ(x) is fixed, while its position and orientation with respect to the external sources may be varied. The situation is analyzed in a body frame of the charge distribution; the origin of that frame should be chosen inside the distribution. Expansion of the external potential around the origin of the body frame yields W = qφ(0) p E(0) 6 q = p = Q ij = ρ(x) d 3 x x ρ(x) d 3 x 3 i,j= Q ij E j x j (0) +... with (3x i x j r 2 δ ij ) ρ(x) d 3 x (5) There, Φ(0), E(0) and E j x j (0) are the potential, the field and the field derivatives due to the external sources at the origin (which is fixed with respect to the charge distribution). The self-energy of the charge distribution is not included in this result. The energy W depends on the location and the orientation of the body and its charge distribution relative to the external field sources. The following applications of Eq. 5 have been discussed: Interaction energy between two electric dipoles. See Eq in textbook. Hyperfine structure (HFS) of atoms. The HFS is due to the interactions between the multipoles of the nuclear charge distribution with the field produced by the electron system at the nucleus. The magnetic-dipole interaction, which presently is not of interest, has been pointed out because it normally dominates the HFS. Electric-dipole and magnetic-quadrupole interactions between nuclei and electron shells are identical zero due to the well-defined parity of the nuclear wavefunction. Therefore, the next-higher HFS term, which we are concerned with in the present context, is the electric-quadrupole interaction. The Casimir formula, derived 936, was quoted (H. B. G. 2

22 Casimir, On the Interaction Between Atomic Nuclei and Electrons, reprinted by W. H. Freeman, San Francisco (963)): W HF S,E2 = 4 E z z eq 0 x=0 3 2C(C + ) 2I(I + )J(J + ) I(2I )J(2J ) with C = F (F + ) I(I + ) J(J + ). (52) There, I, J, and F are the nuclear spin, total electronic spin (orbital plus intrinsic), and hyperfine quantum numbers, respectively. Further, Ez z x=0 is the z-derivative of the z-electric field produced by the electron system at the location of the nucleus, whereby the z-axis is taken parallel to the total electronic spin J. Finally, Q 0 is the nuclear quadrupole moment Q 0 = e (3z 2 r 2 )ρ(x ) d 3 x, calculated in a body frame the z-axis of which coincides with the nuclear symmetry axis (if one models the nucleus as a classical charge distribution). The homework problem 4.6 corresponds to the case F = I + J. Note. To solve the homework problem, the above details are not required. You only need the cylindrical symmetry of the electric field and of the nuclear charge distribution about the z-axis. Electrostatics in dielectric media. In dielectric media, it is desirable to have differential equations for a macroscopic electric field. The macroscopic field is the volume average of the microscopic field, E(x) = E micro (x) volume. The averaging volume is small on a macroscopic scale, but contains many molecules. In contrast to the microscopic field, which is complicated and normally unnecessary to know, the macroscopic field follows simple equations. The microscopically valid (and therefore always valid) homogeneous equation E micro = 0 can be averaged: 0 = E micro volume = E micro volume = E. Thus, it is E = 0, which is the same equation as in free space. We conclude that in dielectric media the electric field will still be derivable from an electric potential Φ(x). The inhomogeneous equation E = ρ/ɛ 0 is modified as follows. The dielectric medium is assumed to contain molecules of species i with corresponding volume densities N i. One molecule of species i carries an average charge q i (which is usually zero) and an average molecular dipole p i. Higher-order multipoles of the molecules are neglected (which is an exceedingly good approximation). We account for free charges (and molecular charges, if 0) in the free-charge density ρ free (x); the subscript is usually omitted and we just write ρ(x). The molecular dipole moments are accounted for via a macroscopic polarization P(x) = i N i(x)p i (x). It was then derived in class that [ɛ 0 E(x) + P(x)] = D(x) = ρ(x), where [ɛ 0 E + P] = D defines the electric displacement field D. Once the polarization P(x) is known, the volume polarization charge ρ pol (x) = P(x) can be calculated. On boundaries of polarized media to the vacuum, there is a surface polarization charge σ pol (x) = ˆn P(x), where the normal vector ˆn is pointing from the medium outward. These polarization charges are not included in the density of free charges ρ(x). In contrast, the variable ρ all in the equation for the microscopic field, E(x) = ρ all ɛ 0, includes both free and polarization charge. Thus, questions 22

23 about charge densities in dielectric media have to be handled very carefully, and a clear distinction between free and polarization charges must be made. Reading and review: Definition of linear and isotropic dielectric media. Definitions of electric permittivity, dielectric constant and electric susceptibility. Boundary conditions at interfaces between dielectric media. Assume fields E and D at the boundary of region and fields E 2 and D 2 at the boundary of region 2. Then, [D 2 (x) D (x)] ˆn = σ(x) [E 2 (x) E (x)] ˆn = 0, (53) with the normal vector ˆn pointing from region to region 2. Thus, the component of D normal to the interface displays a discontinuity of size σ - the surface density of free charges -, while the components of E in plane with the interface are continuous. The validity of these boundary conditions does not require linear and / or isotropic behavior of the dielectric. Examples discussed in class: In infinite volumes with constant permittivity ɛ, the electric field follows the equations E = 0 and E = ρ(x) ɛ. Thus, in electrostatic equations containing ɛ 0 - such as the potentials and fields due to localized charges, the potentials and fields due to charge distributions, capacitances, electrostatic energy, energy of capacitors - the dielectric medium is accounted for by replacing ɛ 0 by the permittivity ɛ = ɛ 0 ɛ r (where ɛ r is the dielectric constant). The polarization charge at a dielectric interface with given E- and D-fields on both sides and with zero free charges has been calculated. The potential Φ(x) due to a polarized object with given polarization P (x) in an otherwise sourceand field-free volume can be obtained by calculation of the polarization charges and subsequent use of a microscopic equation for Φ(x): Φ(x) = 4πɛ 0 ( V \ V P (x ) x x d 3 x + V ˆn P (x ) x x da ), (54) where the volume integral is only over the interior of the polarized object. That is, don t extend the volume integral over the discontinuity of P (x), because that would amount to double-counting the effect of the surface polarization charge. The solution of the following image charge problem has been outlined: The volume z > 0 has permittivity ɛ, and the volume z < 0 has permittivity ɛ 2. A (free) point charge q is located at (0, 0, d) with d > 0. The problem can be solved by assuming two image charges at locations (0, 0, ±d) and consideration of the boundary conditions for D z and for E x, E y on the interface plane z = 0. Reading: Details of this example in Chapter 4.4 of the textbook. 23